1) Consider the following reaction: CO( g )+2H2( g )CH3OH( g ) A reaction mixtur

ID: 882687 • Letter: 1

Question

Consider the following reaction:
CO(g)+2H2(g)CH3OH(g)
A reaction mixture in a 5.17 L flask at a certain temperature initially contains 26.5 g CO and 2.36 g H2. At equilibrium, the flask contains 8.67 gCH3OH.

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

For the following reaction, Kc = 255 at 1000 K.
CO (g) + Cl2 (g) COCl2 (g)
A reaction mixture initially contains a COconcentration of 0.1510 M and a Cl2concentration of 0.178 M at 1000 K.

Part A

What is the equilibrium concentration of CO at 1000 K?

Express your answer in molarity to three significant figures.

Part B

What is the equilibrium concentration of Cl2 at 1000 K?

Express your answer in molarity to three significant figures.

Part C

What is the equilibrium concentration of COCl2 at 1000 K?

Express your answer in molarity to three significant figures.

The reversible chemical reaction

A+BC+D

has the following equilibrium constant:

Kc=[C][D][A][B]=3.0

Part A

Initially, only A and B are present, each at 2.00 M. What is the final concentration ofA once equilibrium is reached?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the final concentration of D at equilibrium if the initial concentrations are[A] = 1.00 M and [B] = 2.00 M ?

Express your answer to two significant figures and include the appropriate units.

The reaction

2CH4(g)C2H2(g)+3H2(g)

has an equilibrium constant of K = 0.154.

If 6.00 mol of CH4, 4.05 mol of C2H2, and 10.10 mol of H2 are added to a reaction vessel with a volume of 5.80 L , what net reaction will occur?

The reaction

has an equilibrium constant of = 0.154.

If 6.00 of , 4.05 of , and 10.10 of are added to a reaction vessel with a volume of 5.80 , what net reaction will occur?

No further reaction will occur because the reaction is at equilibrium.

Part A

For the reaction

2CH4(g)C2H2(g)+3H2(g)

K = 0.150 at 1677 C . What is Kp for the reaction at this temperature?

Express your answer numerically.

Part B

For the reaction

N2(g)+3H2(g)2NH3(g)

Kp = 4.55×103 at 324 C . What is Kfor the reaction at this temperature?

Enter your answer numerically.

Consider the following reaction:
H2 (g) + I2 (g) 2 HI (g)
Complete the following table. Assume that all concentrations are equilibrium concentrations in M.

50.2

Find [H2] at 340 C.

Express your answer using two significant figures.

Find [HI] at 445 C.

Express your answer using three significant figures.

Consider the following reaction:
CO(g)+H2O(g)CO2(g)+H2(g)
Kp=0.0611 at 2000 K
A reaction mixture initially contains a CO partial pressure of 1358 torr and a H2O partial pressure of 1768 torr at 2000 K.

Part A

Calculate the equilibrium partial pressure of CO2

Part B

Calculate the equilibrium partial pressure of H2.

The reaction will proceed to the left to establish equilibrium. The reaction will proceed to the right to establish equilibrium.

No further reaction will occur because the reaction is at equilibrium.

Part A

For the reaction

2CH4(g)C2H2(g)+3H2(g)

K = 0.150 at 1677 C . What is Kp for the reaction at this temperature?

Express your answer numerically.

Part B

For the reaction

N2(g)+3H2(g)2NH3(g)

Kp = 4.55×103 at 324 C . What is Kfor the reaction at this temperature?

Enter your answer numerically.

Consider the following reaction:
H2 (g) + I2 (g) 2 HI (g)
Complete the following table. Assume that all concentrations are equilibrium concentrations in M.

T(C) [H2] [I2] [HI] [Kc] 25 0.0355 0.0388 0.922 340 4.55×102 M 0.388 M 9.6 445 4.85×102M 4.76×102 M

50.2

Find [H2] at 340 C.

Express your answer using two significant figures.

Find [HI] at 445 C.

Express your answer using three significant figures.

Part A

Calculate the equilibrium partial pressure of CO2

Part B

Calculate the equilibrium partial pressure of H2.

Explanation / Answer

1) Given the following reaction:

CO(g)+2H2(g)CH3OH(g)

A reaction mixture in a 5.17 L flask at a certain temperature initially contains 26.5 g CO and 2.36 g H2.

At equilibrium, the flask contains 8.67 g CH3OH.

Total volume = 5.17 L

moles of CH3OH at equlibrium = 8.67/32.04 = 0.2706 mols

Molarity = 0.2706/5.17 = 0.0523 M

Initial moles of CO = 26.5/28.01 = 0.9461 mol

Initial moles of H2 = 2.36/2.02 = 1.17 mol

equilibrium moles of CO = (0.9461-0.2706) = 0.6755 mols

Molarity = 0.6755/5.17 = 0.1307 M

equilibrium moles of H2 = (1.17 - 3x0.2706) = 0.3582 mols

Molarity = 0.3582/5.17 = 0.07 M

Kc = (0.0523)/(0.1307)(0.07)^3 = 1166.63

2) Given for the following reaction, Kc = 255 at 1000 K.

CO (g) + Cl2 (g) COCl2 (g)

A reaction mixture initially contains a CO concentration of 0.1510 M and a Cl2 concentration of 0.178 M at 1000 K.

Part A

The equilibrium concentration of CO at 1000 K

1 M of CO will react with 1 M of Cl2 to give 1 M of COCl2, let x M be the amount that reacted, then,

equilibrium concentration of CO = 0.1510-x M

equilibrium concentration of Cl2 = 0.178-x M

equilibrium concentration of COCl2 = x M

Feed the values in Kc equation,

Kc = 255 = [COCl2]/[CO][Cl2] = x / (0.1510-x)(0.178-x)

Solve for x,

255 = x/(0.0269 - 0.1510x - 0.178x +x^2)

255x^2 - 84.15x + 6.86 = x

255x^2 - 84.9x + 6.9 = 0

x = 0.141

So, the equilibrium concentration of [CO] = 0.1510 - 0.141 = 0.010 M

Part B

the equilibrium concentration of Cl2 at 1000 K = 0.178-0.141 = 0.037 M

Part C

the equilibrium concentration of COCl2 at 1000 K = 0.141 M

3) The reversible chemical reaction

A+BC+D

has the following equilibrium constant:

Kc=[C][D]/[A][B]=3.0

Part A

Initially, only A and B are present, each at 2.00 M.

Calculation for the final concentration of A once equilibrium is reached.

Let x M be reacted, then,

Kc = 3.0 = x^2/(2-x)^2

3.0 = x^2 / (4 - 4x + x^2)

2x^2 - 12x + 12 = 0

x = 1.268 M

So, [A] at equilibrium = 2-1.268 = 0.732 M

Part B

Now the final concentration of D at equilibrium if the initial concentrations are[A] = 1.00 M and [B] = 2.00 M would be,

3.0 = x^2/(1-x)(2-x)

3.0 = x^2/(2 - 3x + x^2)

2x^2 - 9x + 6 = 0

x = 0.814 M

So, the equilibrium concentration of [D] = 0.814 M

4) The reaction

2CH4(g)C2H2(g)+3H2(g)

has an equilibrium constant of K = 0.154.

If 6.00 mol of CH4, 4.05 mol of C2H2, and 10.10 mol of H2 are added to a reaction vessel with a volume of 5.80 L , the net reaction will occur,

The reaction will move towards right to extablish equilibrium

Note the number of moles on left are more, so to establish equilibrium, the reaction would try to equilibriate this condition.

Part A Given, For the reaction

2CH4(g)C2H2(g)+3H2(g)

K = 0.150 at 1677 C .

Kp = Kc(RT)^delta n

delta n = 4-2 = -2

R = gas constant

T = 1677 oC = 1677 + 273 = 1950 K

Kp = 0.150 (160.017)^-2 = 5.86 x 10^-6

Part B

For the reaction

N2(g)+3H2(g)2NH3(g)

Kp = 4.55×103 at 324 C .

Kp = 4.55 x 10^-3 = Kc (0.08206 x 324+273)^(2-4)

Kc = 10.922

7) Given the following reaction:

CO(g)+H2O(g)CO2(g)+H2(g)

Kp=0.0611 at 2000 K

A reaction mixture initially contains a CO partial pressure of 1358 torr and a H2O partial pressure of 1768 torr at 2000 K.

Part A

the equilibrium partial pressure of CO2

Kp = 0.0611 = [CO2][H2]/[CO][H2O] = x^2/(0.1358-x)(0.1768-x)

0.0611 = x^2/(0.024 - 0.313x + x^2)

0.939x^2 + 0.02x - 0.0015 = 0

x = 0.0307

So, equilibrium partial pressure of CO2 = 0.0307 torr

Part B

and the equilibrium partial pressure of H2 = 0.0307 torr

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1) Consider the following reaction: CO( g )+2H2( g )CH3OH( g ) A reaction mixtur

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